Key Fact
Hydrogen bonding
Strong dipole-dipole when H is bonded to F, O, or N. Intermolecular (between molecules) or intramolecular (within molecule). Causes high BP of H₂O, HF, NH₃; protein/DNA structure.
-- NCERT Class 11 Chemistry, Ch. 4, p. 131Hydrogen Bonding
8 MCQs9-step worked example
Source: NCERT Chemical Bonding and Molecular StructureOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Hydrogen Bonding, explained for NEET
Hydrogen bonding is the single most quietly punishing topic in NEET chemical bonding — not because the concept is hard, but because aspirants underestimate its reach. The question rarely says "hydrogen bond." It says "highest boiling point," "maximum viscosity," or "most soluble in water," and the answer turns on whether you correctly identified where hydrogen bonds form and how strong they are.
What qualifies as a hydrogen bond. A hydrogen bond forms when H is covalently bonded to a small, highly electronegative atom — F, O, or N — and the δ⁺ hydrogen interacts with a lone pair on F, O, or N of a neighbouring molecule. This is an intermolecular (or sometimes intramolecular) electrostatic attraction, not a covalent bond. Typical strength: 10–40 kJ/mol, far stronger than van der Waals forces (~1–5 kJ/mol) but weaker than covalent bonds (~150–400 kJ/mol). NCERT Class 11 Chemistry Chapter 4, page 131 explicitly categorises hydrogen bonding as a special case of dipole-dipole interaction arising from the high electronegativity and small size of F, O, and N.
Two types. Intermolecular hydrogen bonding (e.g., H₂O molecules linking to each other) raises boiling point, viscosity, and surface tension. Intramolecular hydrogen bonding (e.g., the –OH and –NO₂ groups within ortho-nitrophenol) forms a closed ring within one molecule, actually lowering the boiling point relative to para-nitrophenol because it reduces intermolecular association.
The NEET trap pattern. Questions comparing boiling points of HF, H₂O, and NH₃ test whether you know that H₂O has the highest boiling point among the three — not HF, despite F being more electronegative — because each water molecule can form four hydrogen bonds (two via H donors, two via O lone pairs), whereas HF forms only two per molecule in a zig-zag chain. The number of hydrogen bonds per molecule matters more than the electronegativity alone.
Watch out: Cl, S, and C are NOT electronegative or small enough to form classical hydrogen bonds. HCl has no hydrogen bonding — its intermolecular force is dipole-dipole only.
Can you answer these Hydrogen Bonding MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following conditions is necessary for the formation of a hydrogen bond?
Show answer and why every option is right or wrong
Answer: B. Hydrogen bonding requires H to be covalently bonded to a small, highly electronegative atom — specifically F, O, or N. Other electronegative atoms like Cl or S are too large and insufficiently electronegative to form classical hydrogen bonds (NCERT Class 11 Chemistry Chapter 4, page 131).
Why A is wrong: A is wrong because not any electronegative atom qualifies — only F, O, and N have the combination of high electronegativity and small atomic size needed. Cl, for instance, is electronegative but too large.
Why C is wrong: C is wrong because metals are electropositive, not electronegative. Metal-H bonds (as in NaH or CaH₂) are ionic or metallic, not the polar covalent bond required for hydrogen bonding.
Why D is wrong: D is wrong because among halogens, only F forms hydrogen bonds. Cl, Br, and I are too large despite being electronegative.
MCQ 2Easy RecallPractice
Hydrogen bonding is a special case of which type of intermolecular force?
Show answer and why every option is right or wrong
Answer: A. NCERT Class 11 Chemistry Chapter 4 classifies hydrogen bonding as a special, particularly strong case of dipole-dipole interaction, arising from the high polarity of the H–F, H–O, or H–N bond.
Why B is wrong: B is wrong because London dispersion forces arise from instantaneous induced dipoles in all molecules, including nonpolar ones. Hydrogen bonding requires a permanent dipole with H bonded to F, O, or N — a fundamentally different origin.
Why C is wrong: C is wrong because ion-dipole interactions involve an ion and a polar molecule (e.g., Na⁺ with H₂O in dissolution). Hydrogen bonding involves two polar molecules, neither of which needs to be an ion.
Why D is wrong: D is wrong because a hydrogen bond is an intermolecular electrostatic attraction (10–40 kJ/mol), not a covalent bond (150–400 kJ/mol). The H remains covalently bonded to its original atom.
MCQ 3Easy RecallPractice
Which of the following molecules does NOT exhibit hydrogen bonding?
Show answer and why every option is right or wrong
Answer: D. HCl does not form hydrogen bonds because Cl, despite being electronegative, is too large in atomic size to create the strong δ⁺H···Cl⁻ interaction that defines hydrogen bonding. Only F, O, and N qualify.
Why A is wrong: A is wrong because H₂O has H bonded to O (one of the three qualifying atoms). Water is the textbook example of extensive hydrogen bonding.
Why B is wrong: B is wrong because HF has H bonded to F, the most electronegative element. HF molecules form hydrogen-bonded zig-zag chains.
Why C is wrong: C is wrong because NH₃ has H bonded to N. Ammonia molecules form hydrogen bonds, which is why NH₃ has an anomalously high boiling point compared to PH₃.
MCQ 4Direct ApplicationPractice
Among HF, H₂O, and NH₃, which has the highest boiling point?
Show answer and why every option is right or wrong
Answer: C. H₂O has the highest boiling point (100 °C vs. 19.5 °C for HF and −33 °C for NH₃) because each water molecule can form four hydrogen bonds — two through its H atoms as donors and two through its O lone pairs as acceptors. HF can form only two per molecule in a linear chain, and NH₃ has only one lone pair available as acceptor despite three N–H donors.
Why A is wrong: A is wrong because higher electronegativity of the partner atom does not automatically mean the highest boiling point. The number of hydrogen bonds per molecule matters more. HF forms only a zig-zag chain (two H-bonds per molecule), while H₂O forms a three-dimensional network (four per molecule).
Why B is wrong: B is wrong because although N is the largest of the three atoms, a larger atom forms weaker hydrogen bonds, not stronger ones. NH₃ has the lowest boiling point of the three.
Why D is wrong: D is wrong because the boiling points differ dramatically — H₂O at 100 °C, HF at 19.5 °C, NH₃ at −33 °C — precisely because the extent of hydrogen bonding differs.
MCQ 5Direct ApplicationPractice
Ortho-nitrophenol has a lower boiling point than para-nitrophenol. The best explanation is:
Show answer and why every option is right or wrong
Answer: D. In ortho-nitrophenol, the –OH and –NO₂ groups are adjacent, allowing an intramolecular hydrogen bond that forms a closed six-membered ring. This "uses up" the H-bonding capacity internally, so fewer intermolecular hydrogen bonds form, leading to less association and a lower boiling point. Para-nitrophenol cannot form such intramolecular bonds (the groups are too far apart), so it relies on extensive intermolecular hydrogen bonding.
Why A is wrong: A is wrong because ortho-nitrophenol does have hydrogen bonding — it is intramolecular. The –OH group still hydrogen-bonds, but to the –NO₂ within the same molecule rather than to neighbouring molecules.
Why B is wrong: B is wrong because both isomers have the same molecular formula (C₆H₅NO₃) and therefore the same molecular mass. The boiling point difference arises entirely from the difference in intermolecular vs. intramolecular hydrogen bonding.
Why C is wrong: C is wrong because para-nitrophenol actually has stronger intermolecular forces (extensive intermolecular hydrogen bonding) than ortho-nitrophenol. That is precisely why para has the higher boiling point.
MCQ 6Direct ApplicationPractice
The typical strength of a hydrogen bond is approximately:
Show answer and why every option is right or wrong
Answer: C. Hydrogen bonds have strengths in the range of approximately 10–40 kJ/mol, placing them between weak van der Waals forces (~1–5 kJ/mol) and strong covalent bonds (~150–400 kJ/mol).
Why A is wrong: A is wrong because 1–5 kJ/mol is the range for London dispersion (van der Waals) forces, which are weaker than hydrogen bonds.
Why B is wrong: B is wrong because 150–400 kJ/mol is the typical range for covalent bond energies. A hydrogen bond is an intermolecular interaction, far weaker than a covalent bond.
Why D is wrong: D is wrong because 400–800 kJ/mol exceeds even most covalent bond energies and corresponds to very strong covalent bonds (e.g., C≡O triple bond). Hydrogen bonds are an order of magnitude weaker.
MCQ 7Concept TrapPractice
Ethanol (C₂H₅OH) is miscible with water in all proportions, while dimethyl ether (CH₃OCH₃) has limited water solubility, despite both having the same molecular formula C₂H₆O. The primary reason is:
Show answer and why every option is right or wrong
Answer: B. Ethanol has an –OH group where H is bonded to O, enabling it to both donate and accept hydrogen bonds with water. Dimethyl ether has no O–H bond; its oxygen can act as a hydrogen bond acceptor but cannot donate, drastically reducing its hydrogen bonding capacity with water and hence its solubility.
Why A is wrong: A is wrong because ethanol and dimethyl ether are structural isomers with the same molecular formula (C₂H₆O) and therefore the same molecular mass. The difference is structural, not mass-related.
Why C is wrong: C is wrong because dimethyl ether is polar (it has a bent C–O–C geometry and a net dipole moment of ~1.3 D). It is not nonpolar. The issue is specifically its inability to donate hydrogen bonds, not a lack of polarity.
Why D is wrong: D is wrong because London dispersion forces depend mainly on molecular size and shape, which are similar for these isomers. The decisive difference is the presence of an –OH hydrogen bond donor in ethanol.
MCQ 8Concept TrapPractice
Ice floats on water because:
Show answer and why every option is right or wrong
Answer: A. In ice, each water molecule forms four hydrogen bonds in a rigid tetrahedral arrangement, creating an open hexagonal lattice with significant empty space. This open structure makes ice less dense than liquid water, where the hydrogen-bonding network is more dynamic and partially collapsed, allowing molecules to pack more closely on average.
Why B is wrong: B is wrong because the O–H covalent bonds within a water molecule do not change in strength between ice and liquid. The difference is in the arrangement of intermolecular hydrogen bonds, not intramolecular covalent bonds.
Why C is wrong: C is wrong because ice actually has more hydrogen bonds per molecule (four per molecule in a fully ordered network) than liquid water (where thermal motion breaks and reforms bonds, averaging fewer at any instant). More hydrogen bonds create the open lattice.
Why D is wrong: D is wrong because freezing is a phase change, not a chemical reaction. The molecular formula and molecular mass of H₂O remain exactly the same in all phases.
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How do you solve a Hydrogen Bonding question? A worked example
- 1
Given
Three Group 15 hydrides: PH₃, NH₃, AsH₃.
- 2
Required
Order of increasing boiling point with reasoning.
- 3
Concept
Boiling point depends on the strength of intermolecular forces. Among these hydrides, only NH₃ can form hydrogen bonds (H bonded to N, which is small and highly electronegative). PH₃ and AsH₃ have only van der Waals (London dispersion) forces because P and As are neither small enough nor electronegative enough to form hydrogen bonds.
- 4
Framework
For PH₃ and AsH₃ (no hydrogen bonding): boiling point increases with molecular mass (stronger London dispersion forces). For NH₃: hydrogen bonding provides an additional, much stronger intermolecular attraction that overrides the mass trend.
- 5
Classification of forces
• PH₃ (M = 34 g/mol): van der Waals only• AsH₃ (M = 78 g/mol): van der Waals only, but higher molecular mass than PH₃• NH₃ (M = 17 g/mol): van der Waals + hydrogen bonding
- 6
Reasoning
Between PH₃ and AsH₃, AsH₃ has the larger electron cloud and higher molecular mass, so its London dispersion forces are stronger → higher boiling point. NH₃, despite having the lowest molecular mass of the three, has hydrogen bonding (10–40 kJ/mol per bond) that far outweighs the weak London forces in PH₃ and AsH₃ (~1–5 kJ/mol).
- 7
Final answer
Increasing boiling point: PH₃ (−87 °C) < AsH₃ (−62.5 °C) < NH₃ (−33 °C).
The expected "mass trend" order would be PH₃ < AsH₃ < NH₃ by mass, but NH₃ is actually the lightest — its anomalously high boiling point is entirely due to hydrogen bonding. - 8
Common trap
Aspirants often predict NH₃ should have the lowest boiling point because it has the lowest molecular mass. This fails because hydrogen bonding in NH₃ dominates over the London forces that govern PH₃ and AsH₃.
- 9
Similar NEET-style question
"Arrange HF, HCl, HBr, HI in order of increasing boiling point." The same logic applies: HCl < HBr < HI follows the mass trend (London forces), but HF is anomalously high due to hydrogen bonding. Answer: HCl < HBr < HI < HF.
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What to remember before solving Hydrogen Bonding questions
More in Chemical Bonding and Molecular Structure: 4 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Hydrogen Bonding questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 18 past-paper questions from Chemical Bonding and Molecular Structure →
Sources
NCERT refs: Class 11 Chemistry Chapter 4, p.131
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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